我需要显示一个货币值的格式1K等于一千,或1.1K, 1.2K, 1.9K等,如果它不是一个偶数千,否则如果低于一千,显示正常500,100,250等,使用JavaScript格式化的数字?
当前回答
改进@tfmontague的答案,进一步格式化小数点。33.0k到33k
largeNumberFormatter(value: number): any {
let result: any = value;
if (value >= 1e3 && value < 1e6) { result = (value / 1e3).toFixed(1).replace(/\.0$/, '') + 'K'; }
if (value >= 1e6 && value < 1e9) { result = (value / 1e6).toFixed(1).replace(/\.0$/, '') + 'M'; }
if (value >= 1e9) { result = (value / 1e9).toFixed(1).replace(/\.0$/, '') + 'T'; }
return result;
}
其他回答
进一步改进@Yash的回答,支持负数:
function nFormatter(num) {
isNegative = false
if (num < 0) {
isNegative = true
}
num = Math.abs(num)
if (num >= 1000000000) {
formattedNumber = (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
} else if (num >= 1000000) {
formattedNumber = (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
} else if (num >= 1000) {
formattedNumber = (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
} else {
formattedNumber = num;
}
if(isNegative) { formattedNumber = '-' + formattedNumber }
return formattedNumber;
}
nFormatter(-120000)
"-120K"
nFormatter(120000)
"120K"
/**
* Shorten number to thousands, millions, billions, etc.
* http://en.wikipedia.org/wiki/Metric_prefix
*
* @param {number} num Number to shorten.
* @param {number} [digits=0] The number of digits to appear after the decimal point.
* @returns {string|number}
*
* @example
* // returns '12.5k'
* shortenLargeNumber(12543, 1)
*
* @example
* // returns '-13k'
* shortenLargeNumber(-12567)
*
* @example
* // returns '51M'
* shortenLargeNumber(51000000)
*
* @example
* // returns 651
* shortenLargeNumber(651)
*
* @example
* // returns 0.12345
* shortenLargeNumber(0.12345)
*/
function shortenLargeNumber(num, digits) {
var units = ['k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'],
decimal;
for(var i=units.length-1; i>=0; i--) {
decimal = Math.pow(1000, i+1);
if(num <= -decimal || num >= decimal) {
return +(num / decimal).toFixed(digits) + units[i];
}
}
return num;
}
谢谢@Cos的评论,我删除了Math。round10依赖。
哇,这里有这么多答案。我想我会告诉你我是如何解决它的,因为它似乎是最容易阅读的,处理负数,并且在JavaScript的千位数范围内。它也很容易改变到你想要的或进一步扩展。
const symbols = [ { value: 1, symbol: '' }, { value: 1e3, symbol: 'k' }, { value: 1e6, symbol: 'M' }, { value: 1e9, symbol: 'G' }, { value: 1e12, symbol: 'T' }, { value: 1e15, symbol: 'P' }, { value: 1e18, symbol: 'E' } ]; function numberFormatter(num, digits) { const numToCheck = Math.abs(num); for (let i = symbols.length - 1; i >= 0; i--) { if (numToCheck >= symbols[i].value) { const newNumber = (num / symbols[i].value).toFixed(digits); return `${newNumber}${symbols[i].symbol}`; } } return '0'; } const tests = [ { num: 1234, digits: 1 }, { num: 100000000, digits: 1 }, { num: 299792458, digits: 1 }, { num: 759878, digits: 1 }, { num: -759878, digits: 0 }, { num: 123, digits: 1 }, { num: 123.456, digits: 1 }, { num: -123.456, digits: 2 }, { num: 123.456, digits: 4 } ]; for (let i = 0; i < tests.length; i++) { console.log(`numberFormatter(${tests[i].num}, ${tests[i].digits})=${numberFormatter(tests[i].num, tests[i].digits)}`); }
通过消除@martin-sznapka解决方案中的循环,您将减少40%的执行时间。
function formatNum(num,digits) {
let units = ['k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'];
let floor = Math.floor(Math.abs(num).toString().length / 3);
let value=+(num / Math.pow(1000, floor))
return value.toFixed(value > 1?digits:2) + units[floor - 1];
}
速度测试(200000随机样本)从这个线程不同的解决方案
Execution time: formatNum 418 ms
Execution time: kFormatter 438 ms it just use "k" no "M".."T"
Execution time: beautify 593 ms doesnt support - negatives
Execution time: shortenLargeNumber 682 ms
Execution time: Intl.NumberFormat 13197ms
我用的是这个函数。它适用于php和javascript。
/**
* @param $n
* @return string
* Use to convert large positive numbers in to short form like 1K+, 100K+, 199K+, 1M+, 10M+, 1B+ etc
*/
function num_format($n) {
$n_format = null;
$suffix = null;
if ($n > 0 && $n < 1000) {
$n_format = Math.floor($n);
$suffix = '';
}
else if ($n == 1000) {
$n_format = Math.floor($n / 1000); //For PHP only use floor function insted of Math.floor()
$suffix = 'K';
}
else if ($n > 1000 && $n < 1000000) {
$n_format = Math.floor($n / 1000);
$suffix = 'K+';
} else if ($n == 1000000) {
$n_format = Math.floor($n / 1000000);
$suffix = 'M';
} else if ($n > 1000000 && $n < 1000000000) {
$n_format = Math.floor($n / 1000000);
$suffix = 'M+';
} else if ($n == 1000000000) {
$n_format = Math.floor($n / 1000000000);
$suffix = 'B';
} else if ($n > 1000000000 && $n < 1000000000000) {
$n_format = Math.floor($n / 1000000000);
$suffix = 'B+';
} else if ($n == 1000000000000) {
$n_format = Math.floor($n / 1000000000000);
$suffix = 'T';
} else if ($n >= 1000000000000) {
$n_format = Math.floor($n / 1000000000000);
$suffix = 'T+';
}
/***** For PHP ******/
// return !empty($n_format . $suffix) ? $n_format . $suffix : 0;
/***** For Javascript ******/
return ($n_format + $suffix).length > 0 ? $n_format + $suffix : 0;
}